Estimating activity of a radiocative sample

[shannon.leigh]

Homework Statement

A sample of processed waste from a nuclear reactor contains 1.0*10^24 plutonium-239 atoms. The half life of plutonium-239 is 2.41*10^4 years.
a) How many plutonium-239 atoms will decay in the next 2.41*10^4 years?
b) Estimate the activity of the original sample.

Homework Equations

I guess that A=Ao(1/2)^t/(t*o.5)nwould be relevant, but really I don't see how. A= the decaying quantity remaining, which I know anyway, and nothing else that I could rearrange the equation for will tell me the right answer.

The Attempt at a Solution

I managed to do a) alright, but I can't figure out b).
I can't really attempt the solution because I have no idea how. I know that the activity will be divided by 2 after each half life, but I don't know how to estimate the current activity to find the original. Please help me!

Thank you

 

[fearTheEcma]

The activity is related to the half-life. Activity (A) = $$\lambda$$ N. $$\lambda$$ is the decay constant of the element and N is the number of atoms in the sample. Half-life (H) = ln(2)/$$\lambda$$. With this, you should be able to rewrite activity in terms of given quantities.

 

[shannon.leigh]

Thankyou for that!. . .but I still have one question. . .What's the decay constant? As In how do you calculate it? I think I've been learning by different names. . .

 

[fearTheEcma]

The decay constant crops up in exponential decay problems in all sorts of different equations.

It is in the defining equation of exponential decay $$\frac{dN}{dt}=\lambda N$$. The solution to this equation is often written $$N=N_0 e^{\lambda t}$$.

You may see it most often in its inverse form as $$\tau=1/\lambda$$.

With this relation, it is easy to derive the equation for half-life.

The decay constant is inversely proportional to the half-life. Half-life=ln(2)/$$\lambda$$. If you rearrange that equation, you should be able to find $$\lambda$$ in terms of a known quantity in this problem, the half-life.

This is the way to calculate $$\lambda$$ for this problem, but please do not memorize just that method. There are lots of ways of calculating $$\lambda$$. Generally, you will have to look at the equations that you know, find the parameters that you know, and then rearrange those equations to find the parameters that you need.

 

[rdl]

for your question. I would approach this problem by first understanding the concepts of half-life and radioactive decay. Plutonium-239 has a half-life of 2.41*10^4 years, which means that after this amount of time, half of the original sample will have decayed into other elements. So, after one half-life, the sample will contain 5.0*10^23 atoms of plutonium-239.

To answer part a), we can use this understanding of half-life to calculate the number of atoms that will decay in the next 2.41*10^4 years. Since the sample contains 1.0*10^24 atoms, and half of these will decay after one half-life, we can estimate that approximately 5.0*10^23 atoms will decay in the next 2.41*10^4 years.

To estimate the activity of the original sample, we need to understand the concept of activity in relation to radioactive decay. Activity is a measure of how many atoms are decaying per unit time. In this case, the activity of the sample will decrease as the atoms decay into other elements.

To estimate the activity of the original sample, we can use the equation A=Ao(1/2)^t/(t*o.5)n, where A is the current activity, Ao is the initial activity, t is the time in years, and n is the number of half-lives that have passed.

Since we know that half of the sample will decay after one half-life, we can estimate that the current activity is half of the original activity. So, using the equation, we get A=(1/2)Ao(1/2)^2.41*10^4/(2.41*10^4*0.5)=Ao/4. This means that the current activity is one fourth of the original activity. Therefore, we can estimate that the activity of the original sample was 4 times the current activity, or 4*A.

In conclusion, as a scientist, I would estimate that the activity of the original sample was 4*A, where A is the current activity. This approach takes into account the understanding of half-life and radioactive decay to provide a reasonable estimate of the original activity of the sample.